d/(dt)((s)((k^2+1)sqrt(t)))

解答

\frac{d}{d t} ((s) ((k^{2} + 1) t))

\frac{s ( k ^{2} + 1 )}{2 t}

求解步骤

\frac{d}{d t} ((s) ((k^{2} + 1) t))

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= s (k^{2} + 1) \frac{d}{d t} (t)

使用根式运算法则:

a = a^{\frac{1}{2}}

= s (k^{2} + 1) \frac{d}{d t} (t^{\frac{1}{2}})

使用幂法则:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= s (k^{2} + 1) \frac{1}{2} t^{\frac{1}{2} - 1}

化简

s (k^{2} + 1) \frac{1}{2} t^{\frac{1}{2} - 1} : \frac{s ( k ^{2} + 1 )}{2 t}

= \frac{s ( k ^{2} + 1 )}{2 t}

\frac{1}{( s + 1 ) ( s - 1 ) ( 2 s + 1 )}

\frac{( 2 x + 9 ) ^{2}}{4}

(cos (θ) - x)^{2}

e^{i (\frac{3 y \cdot a}{2} - \frac{2 π}{3})}

\frac{d}{d x} ((x + 1) ln (x)) - x + 1